LeetCode99 Recover Binary Search Tree 恢复二叉搜索树

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Example 1:

Input: [1,3,null,null,2]

   1
  /
 3
  \
   2

Output: [3,1,null,null,2]

   3
  /
 1
  \
   2

Example 2:

Input: [3,1,4,null,null,2]

  3
 / \
1   4
   /
  2

Output: [2,1,4,null,null,3]

  2
 / \
1   4
   /
  3

Follow up:

• A solution using O(n) space is pretty straight forward.
• Could you devise a constant space solution?

题源：here；完整实现：here

思路：

使用递归的方式遍历二叉树，当遇到逆序的时候就将逆序的值的地址记录下来。为了传递值得地址，我们需要用到指针的引用。代码如下：

void helper(TreeNode* root, int *&val, int *&first, int *&second){
	if (!root) return;

	if (root->left) helper(root->left, val, first, second);

	if (!val) val = &root->val;		
	else if (root->val < *val && !first) {
		first = val; second = &root->val;
	}
	else if (root->val < *val) {
		second = &root->val; return;
	}
	val = &root->val;

	if (root->right) helper(root->right, val, first, second);
}

void recoverTree(TreeNode* root) {
	int *val = NULL, *first = NULL, *second = NULL;
	helper(root, val, first, second);
	int temp = *first;
	*first = *second;
	*second = temp;
}

纪念贴图：